Find Diameter of 2-Cell K in xy Plane

In summary: To prove this, assume the opposite and use the triangle inequality.In summary, the diameter of a set X, as defined by the supremum of the distance function between all possible pairs of points in X, is equal to the diagonal of the rectangle formed by the set, assuming the euclidean metric. This can be proven by finding two points that simultaneously achieve the maximum possible "coordinate" distances and using the triangle inequality to show that this maximizes the euclidean distance.
  • #1
hlin818
30
0

Homework Statement

Define diam(X) of a set X, X a nonempty subset of arbitrary metric space, as diamX=supremum{distancefunction(x,y):x,y in X}.

Let K be the two cell formed by two points a=(2,3) and b=(4,6) in the xy plane. Find diameter of K and show this is the diameter.

Homework Equations


The Attempt at a Solution



my intuition says it is 6, formed by taking the maximum distance around the perimeter of the square. But how would I prove this ( if its even true)? There is also the possibility that diam as defined could be infinite.
 
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  • #2
hlin818 said:

Homework Statement




Define diam(X) of a set X, X a nonempty subset of arbitrary metric space, as diamX=supremum{distance(x,y):x,y in X}.

Let K be the two cell formed by two points a=(2,3) and b=(4,6) in the xy plane. Find diameter of K and show this is the diameter.

Homework Equations





The Attempt at a Solution



my intuition says that the diameter is 6, formed by taking the maximum distance around the perimeter of the square. But how would I prove this ( if its even true)? There is also the possibility that diam as defined could be infinite.

You didn't provide a definition of a "two cell" but that would probably be pertinent to this problem. I don't know what square you're talking about, or how you got the perimeter to be 6. I would be more inclined to say that the diameter is sqrt(13), the distance between the two given points.
 
  • #3
Ah sorry. A 2-cell is a subset of R^2 of the form [a,b] x [c,d], or {(x,y) in R^2: a <= x <= b and c <= y <= d}.

By diameter, since no distance function is specified, how do we know we can't define one that will give a value larger than the straight line path between the two farthest points in the set K?
 
  • #4
So what you are saying is that (2, 3) and (4, 6) are diametrically opposite points in that "2-cell"? They are as far apart as it is possible for points in that 2-cell to be, aren't they? "Distance" between two points isn't measured around a perimeter!
 
  • #5
Yes, that's what I mean.

HallsofIvy said:
So what you are saying is that (2, 3) and (4, 6) are diametrically opposite points in that "2-cell"? They are as far apart as it is possible for points in that 2-cell to be, aren't they? "Distance" between two points isn't measured around a perimeter!

Isn't it true that they are as far apart as possible only in terms of the euclidean distance metric? In finding the diameter, aren't I supposed to find the least upper bound for all possible distance metrics on all possible points in this 2 cell, not just the euclidean distance metric?

What I was trying to get at with the perimeter was that if you measure the maximum distance between two points in the 2 cell along the perimeter, you could have a distance function that for some x,y in K could yield dist(x,y)>the straight line distance.
 
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  • #6
So do the calculation using the euclidean metric. calculate the distance between points on oppoiste of the cell, then use calculus to find the maximum, that will be you diameter.

Mat
 
  • #7
hunt_mat said:
So do the calculation using the euclidean metric. calculate the distance between points on oppoiste of the cell, then use calculus to find the maximum, that will be you diameter.
Mat

I think my main concern is how the euclidean metric yields the largest distance when there are an infinite amount of other possible metrics. How would that account for it?
 
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  • #8
The usual definition of the diameter of a set in a metric space is, as you wrote:

[tex]\sup \{ d(x,y) : x,y \in X\}[/tex]

where, as written, the supremum is taken over varying [itex]x[/itex] and [itex]y[/itex]. The diameter is defined with respect to a specific distance metric. You don't vary the metric; it wouldn't make any sense to do that because then the diameter will always be infinite if the set contains 2 or more points. This is because

[tex]d(x,y) = \begin{cases}
0, & \textrm{if } x = y \\
M, & \textrm{if } x \neq y
\end{cases}[/tex]

is a valid metric for any [itex]M > 0[/itex].
 
  • #9
Ah thanks. Thats precisely what I confused on. I thought diam(K) meant the sup of all possible distance functions so it would have to be infinite, which made no sense.

so I can take it that the question means the euclidean metric?
 
  • #10
hlin818 said:
Ah thanks. Thats precisely what I confused on. I thought diam(K) meant the sup of all possible distance functions so it would have to be infinite, which made no sense.

so I can take it that the question means the euclidean metric?

Probably, unless some other metric is explicitly mentioned. Are there other problems in the same set where the metric is assumed euclidean?
 
  • #11
Generally in this problem set if not specified, you couldn't assume a particular metric. But in any case I think its safe to assume its euclidean. Otherwise the diameter would just be infinite.
 
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  • #12
so how would I start the proof that the diameter is equal to the diagonal of the rectangle formed by the k-cell? it seems so simple but I am having trouble seeing where to begin
 
  • #13
Claim: If you can find two points [itex](x_1, y_1, z_1)[/itex] and [itex](x_2, y_2, z_2)[/itex] which simultaneously achieve the maximum possible "coordinate" distances [itex]|x_2 - x_1|[/itex], [itex]|y_2 - y_1|[/itex], and [itex]|z_2 - z_1|[/itex], then that will maximize the euclidean distance.
 

1. What is a 2-cell K in the xy plane?

A 2-cell K is a geometric shape that can be represented as a polygon with two edges. In the xy plane, it can be visualized as a flat shape with two straight edges connected by two vertices.

2. How do I find the diameter of a 2-cell K in the xy plane?

The diameter of a 2-cell K in the xy plane can be found by measuring the distance between the two vertices of the shape. This distance represents the longest possible straight line that can be drawn within the shape and it is the diameter.

3. Can the diameter of a 2-cell K vary?

Yes, the diameter of a 2-cell K can vary depending on the size and shape of the polygon. However, in the xy plane, the diameter will always be the distance between the two vertices of the shape.

4. How is the diameter of a 2-cell K related to its perimeter?

The diameter of a 2-cell K is not directly related to its perimeter. The perimeter of a shape is the distance around its edges, while the diameter is the distance across the shape. However, in some cases, the diameter may be a part of the perimeter if the shape is a circle or an ellipse.

5. Can the diameter of a 2-cell K be negative?

No, the diameter of a 2-cell K cannot be negative. The diameter is a measure of distance and distance cannot be negative. It will always be a positive value.

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